Question

A 42-g firecracker is at rest at the origin when it explodes into three pieces. The first, with mass 12 g, moves along the x-axis at 35 m/s. The second, with mass 21 g, moves along the y-axis at 29 m/s. Find the velocity of the third piece.

Answer 1

Answer:

u₃  = - 46.66 i  -67.66 j

Explanation:

Given that

M= 42 g

Initial speed ,u= 0 m/s

m₁ = 12 g ,u₁=35 i m/s

m₂=21 g,u₂=29 j m/s

Lets take mass of the third mass = m₃

The speed of the third mass = u₃ m/s

From mass conservation

M= m₁+ m₂ +  m₃

42 = 12 + 21 +  m₃

m₃= 9 g

There is no any external force that is why linear momentum will be conserve

M u = m₁ u₁+m₂u₂+m₃u₃

42 x 0 = 35 i x 12 + 21 j x 29 + 9  u₃

9 u₃  = -420 i - 609 j

u₃  = -46.66 i  -67.66 j

Therefore the speed of the third block will be

u₃  = - 46.66 i  -67.66 j

Answer 2

Answer:

$overrightarrow{v_{3}}=-46.67\widehat{i}-67.67\widehat{j}$

Explanation:

M = 42 g

U = 0

m1 = 12 g

vi = 35 m/s along X axis = 35 i

m2 = 21 g

v2 = 29 m/s along Y axis = 29 j

Let v3 be the velocity of third part. m3 = M - m1 - m2 = 42 g - 12 g - 21 g = 9 g

Use conservation of momentum

$M \times \overrightarrow{U}= m_{1}\overrightarrow{v_{1}}+m_{2}\overrightarrow{v_{2}}+m_{3}\overrightarrow{v_{3}}$

$0=12\times 35\widehat{i}+21\times 29\widehat{j}+9\times\overrightarrow{v_{3}}$

$overrightarrow{v_{3}}=-46.67\widehat{i}-67.67\widehat{j}$